$2lm + 2ln + 4l + 3 = -4m + 5$ Solve for $l$.
Explanation: Combine constant terms on the right. $2lm + 2ln + 4l + {3} = -4m + {5}$ $2lm + 2ln + 4l = -4m + {2}$ Notice that all the terms on the left-hand side of the equation have $l$ in them. $2{l}m + 2{l}n + 4{l} = -4m + 2$ Factor out the $l$ ${l} \cdot \left( 2m + 2n + 4 \right) = -4m + 2$ Isolate the $l$ $l \cdot \left( {2m + 2n + 4} \right) = -4m + 2$ $l = \dfrac{ -4m + 2 }{ {2m + 2n + 4} }$